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I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.

Proof: Notice that $$\langle 1 - i \rangle\\ \Rightarrow 1 = i\\ \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$

This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$

Eric
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    Yes that's correct. Be a bit careful about writing things like $1=i$ when you mean that their images are equivalent in the quotient. It's fine for simple examples but this things can really catch you out later on. – tharris Apr 15 '13 at 23:41
  • Thank you so much. I am not sure if I should delete this post or leave it? – Eric Apr 15 '13 at 23:42
  • I don't know what the protocol is, I'm fairly new here too :) – tharris Apr 15 '13 at 23:43
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    I'd advise you to answer your own question, perhaps make it a community wiki (if you want), and when you are able, mark it accepted. – davidlowryduda Apr 15 '13 at 23:43
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    You also need to prove that $1\not\equiv 0,:$ i.e. that the quotient ring is not trivial. – Math Gems Apr 15 '13 at 23:52
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    This may be a nitpick, but you haven't actually shown that $0 + \langle 1-i \rangle \neq 1 + \langle 1-i \rangle$, so there's a little more work to do. –  Apr 15 '13 at 23:52
  • @Hurkyl No, that's not a nitpick. – Loki Clock Apr 16 '13 at 00:03
  • I wonder if anyone have seen this. –  Apr 16 '13 at 00:06
  • That's a good point, but how do I show that $0 + \langle 1-i \rangle \neq 1 + \langle 1 -i \rangle$? Would I do something like this? If $0 + \langle 1-i \rangle = 1 + \langle 1 -i \rangle$, then $1 = 0$, but we already showed $1=i$, so $0=i$, but $i \not \in \mathbb{Z}$, a contradiction. Right? – Eric Apr 16 '13 at 00:11
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    @Eric You need to show $\rm:1\not\equiv 0,:$ i.e. $\rm:1\not\in (1-{\it i},),:$ i.e. $\rm:1-{\it i},\nmid 1,:$ i.e. $\rm:1-{\it i}\ $ is not a unit. One easy way is to rationalize denominators, as in my answer. Or you can use norms. – Math Gems Apr 16 '13 at 00:24
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    @Eric How do you follow from 1=i to 2=0? – Kenneth.K May 12 '16 at 16:20
  • Possibly related: https://math.stackexchange.com/questions/23358 – Watson Nov 28 '18 at 20:50
  • @Kenneth.K: Set $1-i=0$, and square both, then $1-2i+i^2=2i=2=0$. – MSIS Dec 16 '20 at 01:17

3 Answers3

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Your proof only shows that there are at most two elements. So you also have to check that these two elements differ, i.e. that $1-i$ is not a unit. But instead, you can also do it directly, without any elements at all:

$\mathbb{Z}[i]/(i-1)=\mathbb{Z}[x]/(x^2+1)/(x-1)=\mathbb{Z}/(1^2+1)=\mathbb{F}_2$.

Martin Brandenburg
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Your answer is great, but I'd like to give a different view as well.

A standard first or second example of a Euclidean Domain is the Gaussian integers $\mathbb{Z}[i]$, so that in particular the Gaussian integers form a principal ideal domain. We also know that in PIDs, nonzero prime ideals are maximal. So if we were to show that $1 - i$ is a Gaussian prime, then $\langle 1 - i \rangle$ would be a prime ideal, and thus a maximal ideal. Thus, quotienting by it would give a field.

So how do we show that $1 - i$ is prime? Well, compute its norm (from the Euclidean Domain norm, where $|x + iy| = x^2 + y^2$. Its norm is $2$. Norms are multiplicative, so if $1-i = ab$, then $2 = |a||b|$. But its norm is also an integer, and $2$ is a prime (in the reals). Thus $1-i$ is a prime.

And so we have it.

Community
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davidlowryduda
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One must also prove that the quotient ring is $\ne \{0\}.\:$ Below is a complete proof. $\rm\quad \Bbb Z\stackrel{h}{\to}\, \Bbb Z[{\it i}\,]/(1\!-\!{\it i}\,)\:$ is $\rm\,\color{#0b0}{\bf onto,\:}$ by $\rm\:mod\,\ 1\!-\!{\it i}\,:\ {\it i}\,\equiv 1\phantom{\dfrac{|}{|}}\!\!\!\Rightarrow\:a\!+\!b\,{\it i}\,\equiv a\!+\!b\in \Bbb Z\ $
$\rm\quad n\in ker\ h\iff 1\!-\!{\it i}\,\mid n\iff\phantom{\dfrac{|}{|_|}}\!\!\!\!\!\!\! \dfrac{n}{1\!-\!{\it i}}\, =\, \dfrac{n\,(1\!+\!{\it i}\,)}2\,\in\, \Bbb Z[{\it i}\,] \iff \color{#c00}2\mid n\ $
$\rm\quad So \ \ \ \Bbb Z[{\it i}\,]/(1\!-\!{\it i}\,)\, \color{#0b0}{\bf =\ Im\:h}\,\cong\, \Bbb Z/ker\:h \,=\, \Bbb Z/\color{#c00}2\,\Bbb Z\, =\, \Bbb F_2\ $ $\ \ $ QED

Math Gems
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